Probability of drawing all 4 balls

The first ball can be any of the four with probability $\frac{4}{4}$

The second ball must be any of the other three with probability $\frac{3}{4}$

The third ball must be any of the other two with probability $\frac{2}{4}$

The fourth ball must be the ball that hasn't been selected yet with probability $\frac{1}{4}$

All together,

$$\frac{4}{4}\cdot \frac{3}{4}\cdot \frac{2}{4}\cdot\frac{1}{4}=0.09375$$

We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates.  There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$

The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.

The first draw yields a new ball, guaranteed: $$P(\text{ball 1 new})=1$$

For the second draw, there are $3$ possible new balls and $4$ total balls, so: $$P(\text{ball 2 new})=\frac34$$

For the third, there are $2$ possible new balls and $4$ total balls, so: $$P(\text{ball 3 new})=\frac24=\frac12$$

For the fourth, there is one new ball and there are $4$ total balls, so: $$P(\text{ball 4 new}) = \frac14$$

Thus, the answer is: $$\prod P = 1\cdot\frac34\cdot\frac12\cdot\frac14 = \frac3{32}=0.09375\text{ chance.}$$

The existing solutions provide the correct probability, but do not directly answer the question "What am I doing wrong?"

$(1/4)^4$ is the probability of a specific sequence of draws such as:

red, blue, yellow, orange

blue, yellow, orange, red

yellow, orange, blue, red

The event that "all 4 colors were selected" would occur if any of these sequences occurred. So we must count the number of such sequences (4! = 24) and add up their probabilities, which yields $\displaystyle\frac{4!}{4^4}$